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A Driven Series L-C Circuit
Learning Goal: To understand why a series L-C circuit acts like a short circuit at resonance.
An AC source drives a sinusoidal current of amplitude I_0 and frequency omega into an inductor having inductance L and a capacitor having capacitance C that are connected in series. The current as a function of time is given by I(t) = I_0 \sin (\omega t).
Part A
Recall that the voltages V_L(t) and V_C(t) across the inductor and capacitor are not in phase with the respective currents I_L(t) and I_C(t). In particular, which of the following statements is true for a sinusoidal current driver?
ANSWER:
V_L(t) and V_C(t) both lag their respective currents.
V_L(t) and V_C(t) both lead their respective currents.
V_L(t) lags I_L(t) and V_C(t) leads I_C(t).
V_L(t) leads I_L(t) and V_C(t) lags I_C(t).
The phase angle between voltage and current for inductors and capacitors is 90 degrees, or \frac{\pi}{2} radians. Among other things, this means that no power is dissipated in either the inductor or the capacitor, since the time average of current times voltage, \avg{I(t)V(t)}, is zero.
Part B
What is V(t), the voltage delivered by the current source?
Hint B.1
Current in a series circuit
Note that the current through the current source, the capacitor, and the inductor are all equal at all times.
Part B.2
Find the voltage across the capacitor
What is V_C(t), the voltage across the capacitor as a function of time?
Part B.2.a  

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Part B.2.b  

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Express your answer in terms of I_0, C, omega, and t.
ANSWER:
  V_C(t) =  -\frac{I_{0}}{{\omega}C}{\cos}\left({\omega}t\right) 
Part B.3
What is the voltage across the inductor?
What is V_L(t), the voltage across the inductor as a function of time?
Hint B.3.a  

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Express V_L(t) in terms of I_0, omega, and L.
ANSWER:
  V_L(t) =  I_{0}{\omega}L{\cdot}{\cos}\left({\omega}t\right) 
Hint B.4
Total voltage
You should have defined the voltages across the capacitor and inductor in such a way that the total voltage V_total(t) is equal to V_C(t) + V_L(t), that is, so that total voltage is obtained by adding the voltages at time t, with careful attention paid to the signs.
Express V(t) in terms of some or all of the variables I_0, X_L=\omega L and X_C=\frac{1}{\omega C}, or C and L, omega, and, of course, time t.
ANSWER:
  V(t) =  I_{0}{\cdot}{\cos}\left({\omega}t\right){\cdot}\left(X_{L}-X_{C}\right) 
Part C
With V_L and V_C the amplitudes of the voltages across the inductor and capacitor, which of the following statements is true?
ANSWER:
At very high frequencies V_C \gg V_L and at very low frequencies V_C \ll V_L.
At very high frequencies V_L \gg V_C and at very low frequencies V_L \ll V_C.
V_C \gg V_L for all frequencies.
V_L \gg V_C for all frequencies.
V_C and V_L are about the same at all frequencies.
Part D
The behavior of the L-C circuit provides one example of the phenomenon of resonance. The resonant frequency is \omega_0 = \frac{1}{\sqrt{LC}}. At this frequency, what is the amplitude V_0 of the voltage supplied by the current source?
Express your answer using any or all of the constants given in the problem introduction.
ANSWER:
  V_0 =  0 
Part E
Which of the following statements best explains this fact that at the resonant frequency, there is zero voltage across the capacitor and inductor?
ANSWER:
The voltage V(t) is zero at all times because V_L(t) = -V_C(t).
The voltages V(t); V_L(t); and V_C(t) are zero at all times.
The voltage V(t) is zero only when the current is zero.
The voltage V(t) is zero only at times when the current is stationary (at a max or min).
At the resonant frequency of the circuit, the current source can easily push the current through the series L-C circuit, because the circuit has no voltage drop across it at all! Of course, there is always a voltage across the inductor, and there is always a voltage across the capacitor, since they do have a current passing through them at all times; however, at resonance, these voltages are exactly out of phase, so that the net effect is a current passing through the capacitor and the inductor without any voltage drop at all. The L-C series circuit acts as a short circuit for AC currents exactly at the resonant frequency. For this reason, a series L-C circuit is used as a trap to conduct signals at the resonant frequency to ground.
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